Venus would be an inner planet!
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The net force acting on the airplane is 25N.
Forces acting on the paper airplane when it is in the air:
- The forward force generated by the engine, propeller, or rotor is called thrust. It resists or defeats the drag force. It operates generally perpendicular to the longitudinal axis. However, as will be discussed later, this is not always the case.
- Drag is an airflow disruption generated by the wing, rotor, fuselage, and other projecting surfaces that causes a backward, decelerating force. Drag acts backward and perpendicular to the relative wind, opposing thrust.
- Weight is the total load carried by airplane, including the weight of the crew, fuel, and any cargo or baggage. Due to the influence of gravity, weight pulls the airplane downward.
- Lift—acts perpendicular to the flight path through the center of lift and opposes the weight's downward force. It is produced by the air's dynamic influence on the airfoil.
Given.
Weight of the paper airplane, F1 = 16N
The force of air resistance, F2 = 9N
Net force = F1 + F2
Net force = 25N
Thus, the net force acting on the airplane is 25N.
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Answer:
the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic frictionExplanation:
This exercise uses Newton's second law with the condition that the acceleration is zero, by the time the body begins to slide. At this point the balance of forces is
fr- w || = 0
The expression for friction force is that it is proportional to the coefficient of friction by normal.
fr = μ N
When the system is immobile, the coefficient of friction is called static coefficient and has a value, this is due to the union between the surface, when the movement begins some joints are broken giving rise to coefficient of kinetic friction less than static.
In consequence a lower friction force, which is why the system comes out of balance and begins to accelerate.
μ kinetic <μ static
In all this movement the normal with changed that the angle of the table remains fixed.
Consequently, the reason for the acceleration month that the coefficient of kinetic friction is less than the coefficient of satic friction
Let M = mass of the skier,
v2 = his speed at the end of the track.
By conservation of energy,
1/2 Mv^2 = 1/2 Mv2^2 + Mgh
Dividing by M,
1/2 v^2 = 1/2 v2^2 + gh
Multiplying by 2,
v^2 = v2^2 + 2gh
Or v2^2 = v^2 - 2gh
Or v2^2 = 4.8^2 - 2 * 9.8 * 0.46
Or v2^2 = 23.04 - 9.016
Or v2^2 = 14.024 m^2/s^2-----------------------------(1)
In projectile motion, launch speed = v2
and launch angle theta = 48 deg
Maximum height
H = v2^2 sin^2(theta)/(2g)
Substituting theta = 48 deg and value of v2^2 from (1),
H = 14.024 * sin^2(48 deg)/(2 * 9.8)
Or H = 14.024 * 0.7431^2/19.6
Or H = 14.024 * 0.5523/19.6
Or H = 0.395 m = 0.4 m after rounding off
Ans: 0.4 m
The answer in this question is 0.4 m
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